find the value of k if the sum of the zeros of the polynomial x^2-(k+6)x+2 (2k-1) is half their product
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Answered by
19
sum of roots=(-b)/a
=k+6
and product of roots=c/a
=2(2k-1)
Now according to question sum of roots is equal to half of the product of roots.
So,
=>k+6=1/2 x 2(2k-1)
=>k+6=2k-1
=>(-k)=(-7)
=>k=7
HOPE IT HELPED U
=k+6
and product of roots=c/a
=2(2k-1)
Now according to question sum of roots is equal to half of the product of roots.
So,
=>k+6=1/2 x 2(2k-1)
=>k+6=2k-1
=>(-k)=(-7)
=>k=7
HOPE IT HELPED U
Subham1111:
Thanx bro
for helping me
Answered by
8
Hey there !!
sum of zeroes = (-b)/a = k + 6
product of zeroes = c/a =2(2k-1)
sum of zeroes = half of the product of zeroes
k+6 = 1/2 × 2(2k-1)
k+6 = 1/2 × 4k - 1
k+6 = 2k - 1
6 + 1 = 2k - k
k = 7
sum of zeroes = (-b)/a = k + 6
product of zeroes = c/a =2(2k-1)
sum of zeroes = half of the product of zeroes
k+6 = 1/2 × 2(2k-1)
k+6 = 1/2 × 4k - 1
k+6 = 2k - 1
6 + 1 = 2k - k
k = 7
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